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Q.
If $\theta$ is the acute angle of intersection at a real point of intersection of the circle $x^2 + y^2 = 5$ and the parabola $y ^2= 4x$ then Tan $\theta$ is equal to
Given equations are
$x^{2}+y^{2}=5\,\,\,\,\,\dots(i)$
and $y^{2}=4 x\,\,\,\,\,\dots(ii)$
On solving Eq. (i) and (ii), we get
$x=-5,1$
at $x=-5, y^{2}=-20$ (imaginary value)
$\therefore $ at $x=1, \,y^{2}=4$
$\Rightarrow \,y=\pm 2$
Hence, point of intersection are $(1,2)$ and $(1,-2)$
On differentiating Eq. (i) w. r. t. $x$, we get
$2 x+2 y \frac{d y}{d x}=0 $
$\Rightarrow \, \frac{d y}{d x}=-\frac{x}{y} $
$\therefore \, m_{1}=\left(\frac{d y}{d x}\right)_{(1,2)}=-\frac{1}{2}$
And on differentiating Eq. (ii) w. r. t. $x$, we get
$2 y \frac{d y}{d x}=4$
$\Rightarrow \, \frac{d y}{d x}=\frac{2}{y}$
$m_{2}=\left(\frac{d y}{d x}\right)_{(1,2)}=\frac{2}{2}=1$
Now, $ \tan \,\theta =\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| $
$=\left|\frac{-\frac{1}{2}-1}{1-\frac{1}{2}}\right|=\left|\frac{-\frac{3}{2}}{\frac{1}{2}}\right|=3 $